243 lines
10 KiB
Org Mode
243 lines
10 KiB
Org Mode
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#+TITLE: HW 03
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#+AUTHOR: Elizabeth Hunt
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#+STARTUP: entitiespretty fold inlineimages
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#+LATEX_HEADER: \notindent \notag \usepackage{amsmath} \usepackage[a4paper,margin=1in,portrait]{geometry}
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#+LATEX: \setlength\parindent{0pt}
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#+OPTIONS: toc:nil
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* Question One
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** Three Terms
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\begin{align*}
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Si_3(x) &= \int_0^x \frac{s - \frac{s^3}{3!} + \frac{s^5}{5!}}{s} dx \\
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&= x - \frac{x^3}{(3!)(3)} + \frac{x^5}{(5!)(5)}
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\end{align*}
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** Five Terms
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\begin{align*}
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Si_3(x) &= \int_0^x \frac{s - \frac{s^3}{3!} + \frac{s^5}{5!} - \frac{s^7}{7!} + \frac{s^9}{9!}}{s} dx \\
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&= x - \frac{x^3}{(3!)(3)} + \frac{x^5}{(5!)(5)} - \frac{x^7}{(7!)(7)} + \frac{s^9}{(9!)(9)}
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\end{align*}
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** Ten Terms
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\begin{align*}
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Si_{10}(x) &= \int_0^x \frac{s - \frac{s^3}{3!} + \frac{s^5}{5!} - \frac{s^7}{7!} + \frac{s^9}{9!} - \frac{s^{11}}{11!} + \frac{s^{13}}{13!} - \frac{s^{15}}{15!} + \frac{s^{17}}{17!} - \frac{s^{19}}{19!}}{s} ds \\
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&= x - \frac{x^3}{(3!)(3)} + \frac{x^5}{(5!)(5)} - \frac{x^7}{(7!)(7)} + \frac{s^9}{(9!)(9)} - \frac{s^{11}}{(11!)(11)} + \frac{s^{13}}{(13!)(13)} - \frac{s^{15}}{(15!)(15)} \\
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&+ \frac{s^{17}}{(17!)(17)} - \frac{s^{19}}{(19!)(19)}
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\end{align*}
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* Question Three
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For the second term in the difference quotient, we can expand the taylor series centered at x=a:
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\begin{equation*}
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f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2}(x-a)^2 + \cdots \\
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\end{equation*}
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Which we substitute into the difference quotient:
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\begin{equation*}
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\frac{f(a) - f(a - h)}{h} = \frac{f(a) - (f(a) + f'(a)(x-a) + \frac{f''(a)}{2}(x-a)^2 + \cdots)}{h}
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\end{equation*}
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And subs. $x=a-h$:
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\begin{align*}
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\frac{f(a) - (f(a) + f'(a)(x-a) + \frac{f''(a)}{2}(x-a)^2 + \cdots)}{h} &= -f'(a)(-1) + -\frac{1}{2}f''(a)h \\
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&= f'(a) - \frac{1}{2}f''(a)h + \cdots \\
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\end{align*}
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Which we now plug into the initial $e_{\text{abs}}$:
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\begin{align*}
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e_{\text{abs}} &= |f'(a) - \frac{f(a) - f(a - h)}{h}| \\
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&= |f'(a) - (f'(a) + -\frac{f''(a)}{2}h + \cdots)| \\
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&= |- \frac{1}{2}f''(a)h + \cdots | \\
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\end{align*}
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With the Taylor Remainder theorem we can absorb the series following the second term:
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\begin{equation*}
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e_{\text{abs}} = |- \frac{1}{2}f''(a)h + \cdots | = |\frac{1}{2}f''(\xi)h| \leq Ch
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\end{equation*}
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Thus our error is bounded linearly with $h$.
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* Question Four
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For the first term in the difference quotient we know, from the given notes,
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\begin{equation*}
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f(a+h) = f(a) + f'(a)h + \frac{1}{2}f''(a)h^2 + \frac{1}{6}f'''(a)(h^3)
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\end{equation*}
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And from some of the work in Question Three,
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\begin{equation*}
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f(a - h) = f(a) + f'(a)(-h) + \frac{1}{2}f''(a)(-h)^2 + \frac{1}{6}f'''(a)(-h^3)
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\end{equation*}
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We can substitute immediately into $e_{\text{abs}} = |f'(a) - (\frac{f(a+h) - f(a-h)}{2h})|$:
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\begin{align*}
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e_{\text{abs}} &= |f'(a) - \frac{1}{2h}((f(a) + f'(a)h + \frac{1}{2}f''(a)h^2 + \cdots) - (f(a) - f'(a)h + \frac{1}{2}f''(a)h^2 + \cdots))| \\
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&= |f'(a) - \frac{1}{2h}(2f'(a)h + \frac{1}{6}f'''(a)h^3 + \cdots)| \\
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&= |f'(a) - f'(a) - \frac{1}{12}f'''(a)h^2 + \cdots| \\
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&= |-\frac{1}{12}f'''(a)h^2 + \cdots|
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\end{align*}
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Finally, with the Taylor Remainder theorem we can absorb the series following the third term:
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\begin{equation*}
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e_{\text{abs}} = |-\frac{1}{12}f'''(\xi)h^2| = |\frac{1}{12}f'''(\xi)h^2| \leq Ch^2
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\end{equation*}
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Meaning that as $h$ scales linearly, our error is bounded by $h^2$ as opposed to linearly as in Question Three.
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* Question Six
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** A
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#+BEGIN_SRC lisp
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(load "../lizfcm.asd")
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(ql:quickload :lizfcm)
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(defun f (x)
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(/ (- x 1) (+ x 1)))
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(defun fprime (x)
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(/ 2 (expt (+ x 1) 2)))
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(let ((domain-values (loop for a from 0 to 2
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append
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(loop for i from 0 to 9
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for h = (/ 1.0 (expt 2 i))
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collect (list a h)))))
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(lizfcm.utils:table (:headers '("a" "h" "f'" "\\approx f'" "e_{\\text{abs}}")
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:domain-order (a h)
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:domain-values domain-values)
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(fprime a)
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(lizfcm.approx:fwd-derivative-at 'f a h)
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(abs (- (fprime a)
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(lizfcm.approx:fwd-derivative-at 'f a h)))))
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#+END_SRC
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#+RESULTS:
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| a | h | f' | \approx f' | e_{\text{abs}} |
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| 0 | 1.0 | 2 | 1.0 | 1.0 |
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| 0 | 0.5 | 2 | 1.3333333 | 0.66666675 |
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| 0 | 0.25 | 2 | 1.5999999 | 0.4000001 |
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| 0 | 0.125 | 2 | 1.7777777 | 0.22222233 |
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| 0 | 0.0625 | 2 | 1.8823528 | 0.11764717 |
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| 0 | 0.03125 | 2 | 1.939394 | 0.060606003 |
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| 0 | 0.015625 | 2 | 1.9692307 | 0.030769348 |
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| 0 | 0.0078125 | 2 | 1.9844971 | 0.01550293 |
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| 0 | 0.00390625 | 2 | 1.992218 | 0.0077819824 |
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| 0 | 0.001953125 | 2 | 1.9960938 | 0.00390625 |
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| 1 | 1.0 | 1/2 | 0.33333334 | 0.16666666 |
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| 1 | 0.5 | 1/2 | 0.4 | 0.099999994 |
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| 1 | 0.25 | 1/2 | 0.44444445 | 0.055555552 |
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| 1 | 0.125 | 1/2 | 0.47058824 | 0.029411763 |
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| 1 | 0.0625 | 1/2 | 0.4848485 | 0.015151501 |
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| 1 | 0.03125 | 1/2 | 0.4923077 | 0.0076923072 |
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| 1 | 0.015625 | 1/2 | 0.49612403 | 0.0038759708 |
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| 1 | 0.0078125 | 1/2 | 0.49805447 | 0.0019455254 |
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| 1 | 0.00390625 | 1/2 | 0.49902534 | 0.00097465515 |
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| 1 | 0.001953125 | 1/2 | 0.4995122 | 0.0004878044 |
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| 2 | 1.0 | 2/9 | 0.16666666 | 0.055555567 |
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| 2 | 0.5 | 2/9 | 0.19047618 | 0.031746045 |
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| 2 | 0.25 | 2/9 | 0.2051282 | 0.017094031 |
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| 2 | 0.125 | 2/9 | 0.21333337 | 0.008888856 |
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| 2 | 0.0625 | 2/9 | 0.21768713 | 0.004535094 |
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| 2 | 0.03125 | 2/9 | 0.21993065 | 0.002291575 |
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| 2 | 0.015625 | 2/9 | 0.22106934 | 0.0011528879 |
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| 2 | 0.0078125 | 2/9 | 0.22164536 | 0.00057686865 |
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| 2 | 0.00390625 | 2/9 | 0.22193146 | 0.00029076636 |
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| 2 | 0.001953125 | 2/9 | 0.22207642 | 0.00014580786 |
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* Question Nine
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** C
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#+BEGIN_SRC lisp
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(load "../lizfcm.asd")
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(ql:quickload :lizfcm)
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(defun factorial (n)
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(if (= n 0)
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1
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(* n (factorial (- n 1)))))
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(defun taylor-term (n x)
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(/ (* (expt (- 1) n)
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(expt x (+ (* 2 n) 1)))
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(* (factorial n)
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(+ (* 2 n) 1))))
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(defun f (x &optional (max-iterations 30))
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(let ((sum 0.0))
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(dotimes (n max-iterations)
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(setq sum (+ sum (taylor-term n x))))
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(* sum (/ 2 (sqrt pi)))))
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(defun fprime (x)
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(* (/ 2 (sqrt pi)) (exp (- 0 (* x x)))))
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(let ((domain-values (loop for a from 0 to 1
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append
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(loop for i from 0 to 9
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for h = (/ 1.0 (expt 2 i))
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collect (list a h)))))
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(lizfcm.utils:table (:headers '("a" "h" "f'" "\\approx f'" "e_{\\text{abs}}")
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:domain-order (a h)
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:domain-values domain-values)
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(fprime a)
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(lizfcm.approx:central-derivative-at 'f a h)
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(abs (- (fprime a)
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(lizfcm.approx:central-derivative-at 'f a h)))))
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#+END_SRC
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| a | h | f' | \approx f' | e_{\text{abs}} |
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| 0 | 1.0 | 1.1283791670955126d0 | 0.8427006725464232d0 | 0.28567849454908933d0 |
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| 0 | 0.5 | 1.1283791670955126d0 | 1.0409997446922075d0 | 0.0873794224033051d0 |
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| 0 | 0.25 | 1.1283791670955126d0 | 1.1053055663206806d0 | 0.023073600774832004d0 |
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| 0 | 0.125 | 1.1283791670955126d0 | 1.122529655394656d0 | 0.005849511700856569d0 |
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| 0 | 0.0625 | 1.1283791670955126d0 | 1.1269116944798618d0 | 0.0014674726156507223d0 |
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| 0 | 0.03125 | 1.1283791670955126d0 | 1.1280120131008824d0 | 3.6715399463016496d-4 |
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| 0 | 0.015625 | 1.1283791670955126d0 | 1.1282873617826952d0 | 9.180531281738347d-5 |
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| 0 | 0.0078125 | 1.1283791670955126d0 | 1.128356232581468d0 | 2.293451404455915d-5 |
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| 0 | 0.00390625 | 1.1283791670955126d0 | 1.1283734502811613d0 | 5.71681435124205d-6 |
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| 0 | 0.001953125 | 1.1283791670955126d0 | 1.1283777547060847d0 | 1.4123894278572635d-6 |
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| 1 | 1.0 | 0.41510750774498784d0 | 0.4976611317561498d0 | 0.08255362401116195d0 |
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| 1 | 0.5 | 0.41510750774498784d0 | 0.44560523266293384d0 | 0.030497724917946d0 |
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| 1 | 0.25 | 0.41510750774498784d0 | 0.4234889628937013d0 | 0.008381455148713468d0 |
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| 1 | 0.125 | 0.41510750774498784d0 | 0.41725265825950153d0 | 0.002145150514513694d0 |
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| 1 | 0.0625 | 0.41510750774498784d0 | 0.41564710776310854d0 | 5.396000181207006d-4 |
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| 1 | 0.03125 | 0.41510750774498784d0 | 0.4152414157140871d0 | 1.3390796909928948d-4 |
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| 1 | 0.015625 | 0.41510750774498784d0 | 0.41514241394084905d0 | 3.490619586121735d-5 |
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| 1 | 0.0078125 | 0.41510750774498784d0 | 0.41510582632900395d0 | 1.6814159838896003d-6 |
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| 1 | 0.00390625 | 0.41510750774498784d0 | 0.415092913054238d0 | 1.4594690749825112d-5 |
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| 1 | 0.001953125 | 0.41510750774498784d0 | 0.4150670865046777d0 | 4.0421240310117845d-5 |
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* Question Twelve
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First we'll place a bound on $h$; looking at a graph of $f$ it's pretty obvious from the asymptotes that we don't want to go much further than $|h| = 2 - \frac{pi}{2}$.
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Following similar reasoning as Question Four, we can determine an optimal $h$ by computing $e_{\text{abs}}$ for the central difference, but now including a roundoff error for each time we run $f$
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such that $|f_{\text{machine}}(x) - f(x)| \le \epsilon_{\text{dblprec}}$ (we'll use double precision numbers, from HW 2 we know $\epsilon_{\text{dblprec}} \approx 2.22045 (10^{-16})$).
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We'll just assume $|f_{\text{machine}}(x) - f(x)| = \epsilon_{\text{dblprec}}$ so our new difference quotient becomes:
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\begin{align*}
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e_{\text{abs}} &= |f'(a) - (\frac{f(a+h) - f(a-h) + 2\epsilon_{\text{dblprec}}}{2h})| \\
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&= |\frac{1}{12}f'''(\xi)h^2 + \frac{\epsilon_{\text{dblprec}}}{h}|
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\end{align*}
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Because we bounded our $|h| = 2 - \frac{pi}{2}$ we'll find the maximum value of $f'''$ between $a - (2 - \frac{\pi}{2})$ and $a - (2 - \frac{\pi}{3})$. Using [[https://www.desmos.com/calculator/gen1zpohh2][desmos]] I found this to be -2.
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Thus, $e_{\text{abs}} \leq \frac{1}{6}h^2 + \frac{\epsilon_{\text{dblprec}}}{h}$. Finding the derivative:
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\begin{equation*}
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e' = \frac{1}{3}h - \frac{\epsilon_{\text{dblprec}}}{h^2}
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\end{equation*}
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And solving at $e' = 0$:
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\begin{equation*}
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\frac{1}{3}h = \frac{\epsilon_{\text{dblprec}}}{h^2} \Rightarrow h^3 = 3\epsilon_{\text{dblprec}} \Rightarrow h = (3\epsilon_{\text{dblprec}})^{1/3}
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\end{equation*}
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Which is $\approx (3(2.22045 (10^{-16}))^{\frac{1}{3}} \approx 8.7335 10^{-6}$.
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